Entity-relationship Diagram! Database Trivia Quiz

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| By Sumit Srivastava
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Sumit Srivastava
Community Contributor
Quizzes Created: 2 | Total Attempts: 1,378
Questions: 56 | Attempts: 1,012

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Entity-relationship Diagram! Database Trivia Quiz - Quiz

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Questions and Answers
  • 1. 

    Which of the following gives a logical structure of the database graphically?

    • A.

      Entity-relationship diagram

    • B.

      Entity diagram

    • C.

      Both A and B are correct.

    • D.

      Neither A nor B

    Correct Answer
    A. Entity-relationship diagram
    Explanation
    The correct answer is Entity-relationship diagram. An entity-relationship diagram is a graphical representation of the logical structure of a database. It shows the entities (objects or concepts) in the database, their attributes, and the relationships between them. This diagram helps to visualize the overall structure of the database and how different entities are connected to each other. An entity diagram, on the other hand, may refer to a diagram that represents only the entities in the database without showing the relationships between them. Therefore, the correct answer is Entity-relationship diagram.

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  • 2. 

    Every weak entity set can be converted into a strong entity set by:

    • A.

      Using generalization

    • B.

      Adding appropriate attributes

    • C.

      Using aggregation

    • D.

      None of the above

    Correct Answer
    B. Adding appropriate attributes
    Explanation
    Adding appropriate attributes to a weak entity set can convert it into a strong entity set. Weak entity sets depend on a strong entity set for their existence and cannot exist independently. By adding appropriate attributes, the weak entity set becomes capable of existing independently and no longer relies on another entity set for its existence. This strengthens the entity set and converts it into a strong entity set. Generalization and aggregation are not methods for converting weak entity sets into strong entity sets.

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  • 3. 

    In a one-to-many relationship, the entity that is on the one side of the relationship is called a(n) ________ entity.

    • A.

      Parent

    • B.

      Child

    • C.

      Subtype

    • D.

      Instance

    Correct Answer
    A. Parent
    Explanation
    In a one-to-many relationship, the entity that is on the one side of the relationship is called a Parent entity. This entity is the main entity that has a relationship with multiple instances of another entity, known as the Child entity. The Parent entity acts as the "one" in the relationship, while the Child entity represents the "many" side. The Parent entity typically holds the primary key that is used to establish the relationship with the Child entities.

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  • 4. 

    Weak entity relationship is represented as

    • A.

      Underline

    • B.

      Double line

    • C.

      Double diamond

    • D.

      Double rectangle

    Correct Answer
    C. Double diamond
    Explanation
    A weak entity relationship is represented by a double diamond. A weak entity is an entity that cannot exist without the existence of another entity, known as the identifying entity. The double diamond symbolizes this dependency, indicating that the weak entity relies on the identifying entity for its existence. The double diamond is used to visually represent this relationship in entity-relationship diagrams, making it easier to understand the dependencies between entities.

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  • 5. 

    The ER model includes additional concepts like:

    • A.

      Specialization

    • B.

      Generalization

    • C.

      Categorization

    • D.

      All of the Above

    Correct Answer
    D. All of the Above
    Explanation
    The ER model includes additional concepts like specialization, generalization, and categorization. Specialization is the process of defining subclasses based on a superclass, allowing for more specific attributes and relationships. Generalization is the opposite process, where common attributes and relationships are identified and grouped into a superclass. Categorization is the process of organizing entities into categories or classes based on their common characteristics. Therefore, all of the given options are correct as they represent additional concepts included in the ER model.

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  • 6. 

    E-R modelling techniques is a:

    • A.

      Top-down approach

    • B.

      Bottom-up approach

    • C.

      Left-right approach

    • D.

      None of the above

    Correct Answer
    A. Top-down approach
    Explanation
    E-R modelling techniques is a top-down approach because it starts with identifying the overall structure and then breaks it down into smaller components. In this approach, the focus is on understanding the system as a whole and then gradually refining the details. It involves identifying the entities, their relationships, and attributes, and then creating the entity-relationship diagram accordingly. This approach allows for a systematic and organized way of designing the database by starting from a high-level perspective and gradually moving towards the implementation details.

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  • 7. 

    The total participation by entities is represented in the E-R diagram as

    • A.

      Dashed line

    • B.

      Double line

    • C.

      Double rectangle

    • D.

      Circle

    Correct Answer
    B. Double line
    Explanation
    In an E-R diagram, the double line represents total participation by entities. Total participation means that every entity in one entity set must participate in a relationship with an entity in another entity set. The double line indicates that the participation is mandatory and every entity must be involved in the relationship. This is in contrast to a single line, which represents partial participation, where some entities may choose not to participate in the relationship.

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  • 8. 

    Which relationship is used to represent a specialization entity?

    • A.

      ISA

    • B.

      AIS

    • C.

      ONIS

    • D.

      WHOIS

    Correct Answer
    A. ISA
    Explanation
    The relationship used to represent a specialization entity is ISA. This relationship signifies that one entity is a specialized version of another entity. It is commonly used in database design to represent inheritance and hierarchy in entities.

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  • 9. 

    A _____________ constraint requires that an entity belong to no more than one lower-level entity set.

    • A.

      Disjointness

    • B.

      Uniqueness

    • C.

      Special

    • D.

      Relational

    Correct Answer
    A. Disjointness
    Explanation
    A disjointness constraint requires that an entity belong to no more than one lower-level entity set. This means that an entity cannot simultaneously belong to multiple lower-level entity sets. This constraint ensures that there is no overlap or duplication of entities across different entity sets, maintaining the integrity and uniqueness of the data model.

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  • 10. 

    The completeness constraint may be one of the following: Total generalization or specialization, Partial generalization or specialization. Which is the default?

    • A.

      Total

    • B.

      Partial

    • C.

      Should be specified

    • D.

      Cannot be determined

    Correct Answer
    B. Partial
    Explanation
    The default completeness constraint is partial. This means that if no constraint is specified, it is assumed that the generalization or specialization is only partial, meaning that not all instances of the superclass will have corresponding instances in the subclass.

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  • 11. 

    Which normal form is considered adequate for normal relational database design?

    • A.

      2NF

    • B.

      5NF

    • C.

      4NF

    • D.

      3NF

    Correct Answer
    D. 3NF
    Explanation
    The 3NF (Third Normal Form) is considered adequate for normal relational database design. This normal form ensures that there are no transitive dependencies between non-key attributes in a relation. It helps in reducing data redundancy and improving data integrity by eliminating any fields that are not directly dependent on the primary key. By achieving 3NF, a database design can efficiently organize and store data while minimizing data redundancy and maintaining data consistency.

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  • 12. 

    Consider a schema R(A, B, C, D) and functional dependencies A -> B and C -> D. Then the decomposition of R into R1 (A, B) and R2(C, D) is

    • A.

      Dependency preserving and lossless join

    • B.

      Lossless join but not dependency preserving

    • C.

      Dependency preserving but not lossless join

    • D.

      (d) not dependency preserving and not lossless join

    Correct Answer
    C. Dependency preserving but not lossless join
    Explanation
    The decomposition of R into R1(A, B) and R2(C, D) is dependency preserving because the functional dependencies A -> B and C -> D are preserved in the decomposed relations. However, it is not lossless join because there is no common attribute between R1 and R2 that can be used to join them back together without losing any information.

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  • 13. 

    Relation R with an associated set of functional dependencies, F, is decomposed into BCNF. The redundancy (arising out of functional dependencies) in the resulting set of relations is

    • A.

      Zero

    • B.

      More than zero but less than that of an equivalent 3NF decomposition 

    • C.

      Proportional to the size of F+

    • D.

      Indeterminate

    Correct Answer
    B. More than zero but less than that of an equivalent 3NF decomposition 
    Explanation
    The correct answer is more than zero but less than that of an equivalent 3NF decomposition. When a relation is decomposed into BCNF, it minimizes redundancy by eliminating partial dependencies. However, BCNF does not eliminate all redundancy. There may still be redundancy present due to transitive dependencies. In an equivalent 3NF decomposition, which further eliminates transitive dependencies, the redundancy would be completely eliminated. Therefore, the redundancy in the resulting set of relations from BCNF decomposition is more than zero but less than that of an equivalent 3NF decomposition.

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  • 14. 

    A table has fields F1, F2, F3, F4, and F5, with the following functional dependencies: 1) F1->F3 2) F2->F4 3) (F1,F2)->F5 in terms of normalization, this table is in

    • A.

      1NF

    • B.

      2NF

    • C.

      3NF

    • D.

      NONE OF THE ABOVE

    Correct Answer
    A. 1NF
    Explanation
    This table is in 1NF (First Normal Form) because it satisfies the criteria for 1NF. In 1NF, each attribute in a table must contain only atomic values (indivisible values). The given table has fields F1, F2, F3, F4, and F5, and there are no repeating groups or multivalued attributes. Therefore, it meets the requirements for 1NF.

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  • 15. 

    Consider the following functional dependencies in a database. 1) Date_of_Birth->Age 2) Age->Eligibility 3)Name->Roll_number 4)Roll_number->Name 5)Course_number->Course_name 6)Course_number->Instructor 7)(Roll_number, Course_number)->Grade The relation (Roll_number, Name, Date_of_birth, Age) is

    • A.

      In second normal form but not in third normal form 

    • B.

      In third normal form but not in BCNF 

    • C.

      In BCNF   

    • D.

      In none of the above

    Correct Answer
    D. In none of the above
    Explanation
    The given relation (Roll_number, Name, Date_of_birth, Age) is not in any of the above normal forms because it violates the third normal form and BCNF. The functional dependency Age->Eligibility violates the third normal form because Eligibility is not functionally dependent on the key of the relation. Additionally, the functional dependency Roll_number->Name violates BCNF because it is a partial dependency, where the determinant (Roll_number) does not determine all the attributes in the relation. Therefore, the relation is not in any of the mentioned normal forms.

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  • 16. 

    The relation schema Student_Performance (name, courseNo, rollNo, grade) has the following FDs: 1) name,courseNo->grade 2) rollNo,courseNo->grade 3)name->rollNo 4) rollNo->name The highest normal form of this relation scheme is

    • A.

      2NF

    • B.

      BCNF

    • C.

      3NF

    • D.

      4NF

    Correct Answer
    C. 3NF
    Explanation
    The highest normal form of this relation schema is 3NF. This is because all the functional dependencies in the schema are either full key dependencies or transitive dependencies. In 3NF, there should be no transitive dependencies, meaning that no non-key attribute should be functionally dependent on another non-key attribute. In this schema, all the functional dependencies are either between the key attributes (name, courseNo, and rollNo) and the non-key attribute (grade), or between the non-key attributes (name and rollNo). Therefore, the schema satisfies the requirements of 3NF.

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  • 17. 

    The relation EMPDT1 is defined with attributes empcode(unique), name, street, city, state, and pincode. For any pincode,there is only one city and state. Also, for any given street, city and state, there is just one pincode. In normalization terms EMPDT1 is a relation in

    • A.

      1NF only

    • B.

      3NF and hence also in 2NF and 1NF 

    • C.

      BCNF and hence also in 3NF, 2NF and 1NF

    • D.

      2NF and hence also in 1NF

    Correct Answer
    D. 2NF and hence also in 1NF
    Explanation
    The given relation EMPDT1 satisfies the conditions of 2NF and 1NF. In 2NF, there should be no partial dependencies, meaning that no non-key attribute depends on only a part of the primary key. Since for any given street, city, and state, there is just one pincode, there is no partial dependency. In 1NF, each attribute should be atomic and single-valued, which is satisfied by the given relation. Therefore, the relation EMPDT1 is in 2NF and hence also in 1NF.

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  • 18. 

    Which one of the following statements if FALSE?

    • A.

      A prime attribute can be transitively dependent on a key in a BCNF relation.

    • B.

      Any relation with two attributes is in BCNF

    • C.

      A relation in which every key has only one attribute is in 2NF

    • D.

      A prime attribute can be transitively dependent on a key in a 3 NF relation.

    Correct Answer
    A. A prime attribute can be transitively dependent on a key in a BCNF relation.
    Explanation
    In a BCNF (Boyce-Codd Normal Form) relation, a prime attribute cannot be transitively dependent on a key. BCNF requires that every non-trivial functional dependency in the relation must be determined by a superkey. Therefore, a prime attribute cannot be transitively dependent on a key in a BCNF relation.

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  • 19. 

    Consider the following relational schemes for a library database: Book (Title, Author, Catalog_no, Publisher, Year, Price) Collection (Title, Author, Catalog_no) With the following functional dependencies: I. Title Author -> Catalog_no II. Catalog_no -> Title Author Publisher Year III. Publisher Title Year -> Price Assume {Author, Title} is the key for both schemes. Which of the following statements is true?

    • A.

      Both Book and Collection are in 3NF only

    • B.

      Book is in 2NF and Collection is in 3NF

    • C.

      Both Book and Collection are in BCNF

    • D.

      Both Book and Collection are in 2NF only

    Correct Answer
    B. Book is in 2NF and Collection is in 3NF
    Explanation
    The given answer is correct because in the Book scheme, the functional dependency I violates the 2NF as it is a partial dependency. By decomposing the scheme, we can remove this partial dependency and achieve 2NF. In the Collection scheme, there are no partial dependencies or transitive dependencies, so it is already in 3NF. Therefore, the statement "Book is in 2NF and Collection is in 3NF" is true.

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  • 20. 

    Let R(A,B,C,D,E,P,G) be a relational schema in which the following FDs are known to hold: 1) AB->CD 2) DE->P 3) C->E 4) P->C 5) B->G The relation schema R is

    • A.

      Not in 2NF

    • B.

      In 3NF, but not in BCNF

    • C.

      In 2NF, but not in 3NF      

    • D.

      In BCNF  

    Correct Answer
    A. Not in 2NF
    Explanation
    The given relational schema is not in 2NF because it violates the second normal form (2NF) rule which states that no non-prime attribute should be functionally dependent on a proper subset of any candidate key. In this case, attribute E is functionally dependent on C, which is a proper subset of the candidate key AB. Therefore, the relational schema is not in 2NF.

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  • 21. 

    Relation R has eight attributes ABCDEFGH. Fields of R contain only atomic values. F={CH→G, A→BC, B→CFH, E→A, F→EG} is a set of functional dependencies (FDs) so that F + is exactly the set of FDs that hold for R. How many candidate keys does the relation R have?

    • A.

      3

    • B.

      5

    • C.

      4

    • D.

      6

    Correct Answer
    C. 4
    Explanation
    The relation R has four candidate keys. A candidate key is a minimal set of attributes that can uniquely identify every tuple in a relation. To determine the candidate keys, we can use the closure of each attribute set in R. Starting with each individual attribute, we find the closure by applying the functional dependencies in F. The closure of each attribute set is as follows: A+ = ABCFHG, B+ = BCFHGA, C+ = CFHGA, D+ = D, E+ = AEBCFHG, F+ = FEG, G+ = G, H+ = H. From these closures, we can see that the candidate keys are AD, AE, AF, and AG, as they are the minimal sets of attributes that can determine all other attributes in R. Therefore, the answer is 4.

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  • 22. 

    Relation R has eight attributes ABCDEFGH. Fields of R contain only atomic values. F={CH→G, A→BC, B→CFH, E→A, F→EG} is a set of functional dependencies (FDs) so that F + is exactly the set of FDs that hold for R. The relation R is

    • A.

      In 2NF, but not in 3NF. 

    • B.

      In 3NF, but not in BCNF. 

    • C.

      In 1NF, but not in 2NF. 

    • D.

      In BCNF.

    Correct Answer
    C. In 1NF, but not in 2NF. 
    Explanation
    The given set of functional dependencies (FDs) F={CH→G, A→BC, B→CFH, E→A, F→EG} indicates that there are partial dependencies present in the relation R. This means that there are attributes that are functionally dependent on only a part of the candidate key. Therefore, the relation R is in 1NF, but not in 2NF. In 2NF, all non-key attributes should be fully dependent on the entire candidate key. However, the given FDs show that attribute C is dependent on only a part of the candidate key (A), violating the 2NF criteria.

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  • 23. 

    Every time attribute A appears, it is matched with the same value of attribute B, but not the same value of attribute C. Therefore, it is true that:

    • A.

      A → B.

    • B.

      A → C.

    • C.

      A → (B,C).

    • D.

      (B,C) → A.

    Correct Answer
    A. A → B.
    Explanation
    The given information states that every time attribute A appears, it is matched with the same value of attribute B, but not the same value of attribute C. This implies that there is a direct relationship between A and B, where A determines the value of B. However, there is no information provided about the relationship between A and C or between B and C. Therefore, the only valid conclusion that can be drawn is that A determines the value of B, making the statement A → B true.

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  • 24. 

    A functional dependency is a relationship between or among:

    • A.

      Tables

    • B.

      Rows

    • C.

      Relations

    • D.

      Attributes

    Correct Answer
    D. Attributes
    Explanation
    A functional dependency is a relationship between attributes in a database. It states that the value of one or more attributes determines the value of another attribute. In other words, if we know the value of certain attributes, we can determine the value of other attributes based on this relationship. Therefore, the correct answer is "Attributes" because functional dependencies are defined between attributes, not tables, rows, or relations.

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  • 25. 

    If every functional dependency in set E is also in closure of F then this is classified as

    • A.

      FD is covered by E

    • B.

      E is covered by F

    • C.

      F is covered by E

    • D.

      F+ is covered by E

    Correct Answer
    B. E is covered by F
    Explanation
    If every functional dependency in set E is also in the closure of F, it means that the closure of F contains all the functional dependencies in E. This implies that F is able to derive all the functional dependencies in E. Therefore, E is covered by F.

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  • 26. 

    Inst_dept (ID, name, salary, dept name, building, budget) is decomposed into instructor (ID, name, dept name, salary) department (dept name, building, budget) This comes under

    • A.

      Lossy-join decomposition

    • B.

      Lossy decomposition

    • C.

      Lossless-join decomposition

    • D.

      Both a and b

    Correct Answer
    D. Both a and b
    Explanation
    This decomposition can be considered both a lossy-join decomposition and a lossless-join decomposition. It is lossy because the original relation Inst_dept cannot be recreated from the decomposed relations instructor and department. Information about the instructor's salary is lost in the decomposition. However, it is also lossless because the join of the decomposed relations instructor and department can be used to recreate the original relation Inst_dept. Therefore, the correct answer is both a and b.

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  • 27. 

    Consider a relation R(A,B,C,D,E) with the following functional dependencies: ABC -> DE and D -> AB The number of superkeys of R is:

    • A.

      2

    • B.

      7

    • C.

      10

    • D.

      15

    Correct Answer
    C. 10
    Explanation
    The given relation R has two functional dependencies: ABC -> DE and D -> AB. To find the number of superkeys, we need to consider all possible combinations of attributes that can uniquely determine all other attributes.

    From the given functional dependencies, we can see that ABCD is a superkey since it can determine all attributes. Additionally, D is also a superkey as it can determine AB, and ABC is also a superkey as it can determine DE.

    Other possible superkeys can be formed by combining attributes from ABCD with attributes from DE. For example, ABCDE is a superkey since it can determine all attributes. Similarly, ABCE, ACDE, BCDE, ABD, ACD, BCD, and ACE are also superkeys.

    Therefore, the number of superkeys for relation R is 10.

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  • 28. 

    In a two-phase locking protocol, a transaction release locks in ......... phase.

    • A.

      Shrinking phase

    • B.

      Growing phase

    • C.

      Running phase

    • D.

      Initial phase

    Correct Answer
    A. Shrinking pHase
    Explanation
    In a two-phase locking protocol, the transaction releases locks in the shrinking phase. This phase occurs after the transaction has completed its operations and is ready to release the locks it acquired during the growing phase. During the shrinking phase, the transaction releases the locks in a systematic manner, ensuring that other transactions can access the locked resources. This helps in maintaining data consistency and preventing conflicts between concurrent transactions.

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  • 29. 

    ............ protocol grantees that a set of transactions becomes serializable.

    • A.

      Two phase locking

    • B.

      Two phase commit

    • C.

      Transaction locking

    • D.

      Checkpoints

    Correct Answer
    A. Two pHase locking
    Explanation
    Two phase locking is a protocol that ensures serializability of transactions. It consists of two phases: the growing phase and the shrinking phase. In the growing phase, transactions acquire locks on the resources they need before accessing them. This ensures that no other transaction can access the same resource simultaneously. In the shrinking phase, transactions release the locks they acquired, allowing other transactions to access the resources. By following this protocol, the transactions are executed in a serial order, preventing conflicts and ensuring serializability.

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  • 30. 

    The situation in which a transaction holds a data item and waits for the release of data item held by some other transaction, which in turn waits for another transaction, is called .......

    • A.

      Serialiable schedule

    • B.

      Process waiting

    • C.

      Concurrency

    • D.

      Deadlock

    Correct Answer
    D. Deadlock
    Explanation
    Deadlock refers to a situation where multiple transactions are waiting indefinitely for each other to release the resources they hold. In this scenario, a transaction holds a data item and is waiting for the release of the data item held by another transaction. However, that transaction is also waiting for yet another transaction, creating a cycle of dependencies and preventing any progress. This deadlock situation can cause a system to become unresponsive and requires intervention to resolve.

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  • 31. 

    The database system must take special actions to ensure that transactions operate properly without interference from concurrently executing database statements. This property is referred to as

    • A.

      Atomicity

    • B.

      Durability

    • C.

      Isolation

    • D.

      Consistency

    Correct Answer
    C. Isolation
    Explanation
    The given correct answer is "Isolation." Isolation refers to the property of a database system that ensures that transactions are executed in a way that they are not affected by other concurrently executing transactions. It ensures that each transaction is executed independently and in isolation from other transactions, preventing interference and maintaining data integrity. Isolation is important to prevent issues like dirty reads, non-repeatable reads, and phantom reads, which can occur when multiple transactions access and modify the same data simultaneously.

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  • 32. 

    Which of the following is a procedure for acquiring the necessary locks for a transaction where all necessary locks are acquired before any are released?

    • A.

      Record controller

    • B.

      Exclusive lock

    • C.

      Authorization rule

    • D.

      Two phase lock

    Correct Answer
    D. Two pHase lock
    Explanation
    The Two Phase Lock (2PL) protocol is a procedure for acquiring locks in a transaction where all necessary locks are acquired before any are released. In the 2PL protocol, a transaction is divided into two phases: the growing phase and the shrinking phase. In the growing phase, the transaction acquires all the necessary locks before it starts accessing and modifying the data. Once the transaction has acquired all the necessary locks, it enters the shrinking phase where it starts releasing the locks only after it has completed all its data access and modification operations. This ensures that no conflicts occur between transactions and maintains data integrity.

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  • 33. 

    When transaction Ti requests a data item currently held by Tj , Ti is allowed to wait only if it has a timestamp smaller than that of Tj (that is, Ti is older than Tj ). Otherwise, Ti is rolled back (dies). This is

    • A.

      Wait-die

    • B.

      Wait-wound

    • C.

      Wound-wait

    • D.

      Wait

    Correct Answer
    A. Wait-die
    Explanation
    The given scenario describes the "Wait-die" strategy for handling conflicting transactions. In this strategy, if a transaction Ti requests a data item currently held by Tj, Ti is allowed to wait only if it has a smaller timestamp (i.e., it is older) than Tj. If Ti has a larger timestamp (i.e., it is younger) than Tj, Ti is rolled back (dies). This strategy ensures that older transactions are given priority and prevents deadlock situations.

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  • 34. 

    Immediate database modification technique uses

    • A.

      Both undo and redo

    • B.

      Undo but no redo

    • C.

      Redo but no undo

    • D.

      Neither undo nor redo

    Correct Answer
    A. Both undo and redo
    Explanation
    Immediate database modification technique uses both undo and redo. Undo is used to rollback or undo the changes made to the database in case of a failure or error. Redo is used to reapply the changes made to the database after a failure or error has been corrected. By using both undo and redo, the database can maintain consistency and integrity even in the event of failures or errors.

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  • 35. 

    Consider the following transactions with data items P and Q initialized to zero: T1: read (P) ; read (Q) ; if P = 0 then Q : = Q + 1 ; write (Q) ; T2: read (Q) ; read (P) ; if Q = 0 then P : = P + 1 ; write (P) ; Any non-serial interleaving of T1 and T2 for concurrent execution leads to..

    • A.

      A Serializable Schedule

    • B.

      A Schedule that is not conflict Serializable

    • C.

      A Conflict Serializable Schedule

    • D.

      A Schedule for which a precedence graph cannot be drawn

    Correct Answer
    B. A Schedule that is not conflict Serializable
    Explanation
    A non-serial interleaving of T1 and T2 for concurrent execution leads to a schedule that is not conflict serializable. Conflict serializability is a property of a schedule where the order of conflicting operations (operations that access the same data item) is preserved. In this case, both T1 and T2 read and write the same data items P and Q, and their operations conflict with each other. If the schedule is not conflict serializable, it means that the order of conflicting operations is not preserved, leading to potential data inconsistencies.

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  • 36. 

    Consider the transactions T1, T2, and T3 and the schedules S1 and S2 given below. T1: r1(X); r1(Z); w1(X); w1(Z) T2: r2(Y); r2(Z); w2(Z) T3: r3(Y); r3(X); w3(Y) S1: r1(X); r3(Y); r3(X); r2(Y); r2(Z); w3(Y); w2(Z); r1(Z); w1(X); w1(Z) S2: r1(X); r3(Y); r2(Y); r3(X); r1(Z); r2(Z); w3(Y); w1(X); w2(Z); w1(Z) Which one of the following statements about the schedules is TRUE?

    • A.

      Only S1 is conflict Serializable

    • B.

      Only S2 is conflict Serializable

    • C.

      Both S1 and S2 is conflict Serializable

    • D.

      Neither S1 nor S2 is conflict Serializable

    Correct Answer
    A. Only S1 is conflict Serializable
    Explanation
    The schedule S1 is conflict serializable because it can be transformed into an equivalent serial schedule by swapping non-conflicting operations. On the other hand, the schedule S2 is not conflict serializable because there is a conflict between the operations r1(Z) and w2(Z), which cannot be resolved by swapping operations. Therefore, the statement "Only S1 is conflict Serializable" is true.

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  • 37. 

    Consider the following transaction involving two bank accounts x and y. read(x); x := x – 50; write(x); read(y); y := y +50; write(y) The constraint that the sum of the accounts x and y should remain constant is that of

    • A.

      Atomicity

    • B.

      Consistency

    • C.

      Durability

    • D.

      Isolation

    Correct Answer
    B. Consistency
    Explanation
    The constraint that the sum of the accounts x and y should remain constant is that of consistency. Consistency ensures that the data in the database remains in a valid state before and after the transaction. In this case, the transaction involves deducting 50 from account x and adding 50 to account y. To maintain consistency, the sum of the balances in both accounts should remain the same before and after the transaction.

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  • 38. 

    Which level of RAID refers to disk mirroring with block striping?

    • A.

      RAID level 1

    • B.

      RAID level 2

    • C.

      RAID level 3

    • D.

      RAID level 4

    Correct Answer
    A. RAID level 1
    Explanation
    RAID level 1 refers to disk mirroring with block striping. In this level of RAID, data is mirrored onto two or more disks simultaneously, providing redundancy and fault tolerance. Each disk in the array contains an exact copy of the data, allowing for data recovery in case of disk failure. Block striping is also used, which means that data is divided into blocks and distributed across the disks in the array. This combination of mirroring and striping provides both data protection and improved performance.

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  • 39. 

    A unit of storage that can store one or more records in a hash file organization is denoted as

    • A.

      Buckets

    • B.

      Disk Pages

    • C.

      Blocks

    • D.

      Nodes

    Correct Answer
    A. Buckets
    Explanation
    A unit of storage that can store one or more records in a hash file organization is denoted as "Buckets". In a hash file organization, data is stored in buckets, which are essentially containers that hold multiple records. Each bucket is identified by a unique hash value, allowing for efficient retrieval and storage of records based on their key values. Buckets are commonly used in hash table data structures to store and retrieve data quickly using a hashing algorithm.

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  • 40. 

    The file organization which allows us to read records that would satisfy the join condition by using one block read is

    • A.

      Heap file organization

    • B.

      Sequential file organization

    • C.

      Clustering file organization

    • D.

      Hash file organization

    Correct Answer
    C. Clustering file organization
    Explanation
    Clustering file organization allows us to read records that would satisfy the join condition by using one block read. In this type of file organization, related records are physically stored together in the same block or nearby blocks on the disk. This reduces the number of disk accesses required to retrieve the records needed for a join operation, resulting in improved performance.

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  • 41. 

    The highest level in the hierarchy of data organization is called

    • A.

      Databank

    • B.

      Database

    • C.

      Datafile

    • D.

      Datarecord

    Correct Answer
    B. Database
    Explanation
    The highest level in the hierarchy of data organization is called a database. A database is a collection of related data that is organized and stored in a structured manner. It allows for efficient storage, retrieval, and manipulation of data. At the highest level, a database contains multiple tables or files, which in turn contain records and fields. It provides a centralized and integrated approach to managing data, ensuring data consistency and security.

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  • 42. 

    Which of the following hardware component is the most important to the operation of database management system?

    • A.

      High Resolution Video Display

    • B.

      Printer

    • C.

      High Speed Large Capacity Disk

    • D.

      Mouse

    Correct Answer
    C. High Speed Large Capacity Disk
    Explanation
    The most important hardware component for the operation of a database management system is a high speed large capacity disk. This is because a database management system requires a large amount of storage space to store and retrieve data efficiently. A high speed disk ensures fast access and retrieval of data, while a large capacity disk allows for the storage of a large amount of data. Without a high speed large capacity disk, the performance of the database management system would be significantly hindered.

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  • 43. 

    What operator tests column for the absence of data?

    • A.

      EXISTS operator

    • B.

      NOT operator

    • C.

      IS NULL operator

    • D.

      None of these

    Correct Answer
    C. IS NULL operator
    Explanation
    The IS NULL operator tests a column for the absence of data. It is used to check if a column contains a NULL value, which represents the absence of data. Therefore, the IS NULL operator is the correct answer for this question.

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  • 44. 

    In SQL, which command(s) is(are) used to change a table's storage characteristics?

    • A.

      ALTER TABLE

    • B.

      MODIFY TABLE

    • C.

      CHANGE TABLE

    • D.

      All of the above

    Correct Answer
    A. ALTER TABLE
    Explanation
    The ALTER TABLE command is used in SQL to change a table's storage characteristics. This command allows modifications to be made to an existing table, such as adding or dropping columns, changing data types, or altering constraints. The MODIFY TABLE and CHANGE TABLE options are not valid SQL commands for changing a table's storage characteristics. Therefore, the correct answer is ALTER TABLE.

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  • 45. 

    Which of the following statement on the view concept in SQL is invalid ?

    • A.

      All views are not updatable

    • B.

      The views may be referenced in an SQL statement whenever tables are referenced

    • C.

      The views are instantiated at the time they are referenced and not when they are defined

    • D.

      The definition of a view should not have GROUP BY clause in it

    Correct Answer
    D. The definition of a view should not have GROUP BY clause in it
    Explanation
    The given correct answer is "the definition of a view should not have GROUP BY clause in it". This statement is invalid because a view can have a GROUP BY clause in its definition. The GROUP BY clause is used to group rows based on a specific column or columns in the view's underlying query. It is often used in conjunction with aggregate functions to perform calculations on grouped data. Therefore, the statement is incorrect.

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  • 46. 

    Find the temperature in increasing order of all cities

    • A.

      SELECT city FROM weather ORDER BY temperature;

    • B.

      SELECT city, temperature FROM weather;

    • C.

      SELECT city, temperature FROM weather ORDER BY temperature;

    • D.

      SELECT city, temperature FROM weather ORDER BY city;

    Correct Answer
    C. SELECT city, temperature FROM weather ORDER BY temperature;
    Explanation
    This answer is correct because it selects both the city and temperature columns from the weather table and orders the results in increasing order based on the temperature column. This will give a list of cities sorted by their temperature values from lowest to highest.

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  • 47. 

    Find the names of these cities with temperature and condition whose condition is neither sunny nor cloudy

    • A.

      SELECT city, temperature, condition FROM weather WHERE condition NOT IN ('sunny', 'cloudy');

    • B.

      SELECT city, temperature, condition FROM weather WHERE condition NOT BETWEEN ('sunny', 'cloudy');

    • C.

      SELECT city, temperature, condition FROM weather WHERE condition IN ('sunny', 'cloudy');

    • D.

      SELECT city, temperature, condition FROM weather WHERE condition BETWEEN ('sunny', 'cloudy');

    Correct Answer
    A. SELECT city, temperature, condition FROM weather WHERE condition NOT IN ('sunny', 'cloudy');
    Explanation
    The correct answer is the first option: SELECT city, temperature, condition FROM weather WHERE condition NOT IN ('sunny', 'cloudy'). This query retrieves the names of cities along with their temperature and condition from the weather table. The condition is filtered to exclude both 'sunny' and 'cloudy', meaning that only cities with conditions other than sunny or cloudy will be included in the result.

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  • 48. 

    Which of the following is not true of the traditional approach to information processing

    • A.

      There is common sharing of data among the various applications

    • B.

      It is file oriented

    • C.

      Programs are dependent on the file

    • D.

      It is inflexible

    Correct Answer
    A. There is common sharing of data among the various applications
    Explanation
    The traditional approach to information processing does not involve common sharing of data among various applications. In this approach, each application has its own separate data files and does not share data with other applications. This lack of data sharing can lead to data redundancy and inconsistency.

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  • 49. 

    The information about data in a database is called _______.

    • A.

      Metadata

    • B.

      Hyperdata

    • C.

      Teradata

    • D.

      None of these

    Correct Answer
    A. Metadata
    Explanation
    Metadata refers to the information about data in a database. It includes details such as the structure, format, and organization of the data, as well as information about its source, meaning, and usage. Metadata provides context and helps in understanding and managing the data effectively. It is used for data integration, data governance, data quality, and other data management processes. Hyperdata and Teradata are not valid terms related to the information about data in a database. Therefore, the correct answer is Metadata.

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  • 50. 

    A data dictionary is a special file that contains?

    • A.

      The names of all fields in all files

    • B.

      The data types of all fields in all files

    • C.

      The widths of all fields in all files

    • D.

      All of the mentioned

    Correct Answer
    D. All of the mentioned
    Explanation
    A data dictionary is a special file that contains all the mentioned information. It includes the names of all fields in all files, the data types of all fields in all files, and the widths of all fields in all files. The data dictionary serves as a reference guide for understanding the structure and organization of the data stored in a database or system. It helps in ensuring consistency and accuracy in data management by providing a centralized source of information about the data elements used in the system.

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Quiz Review Timeline +

Our quizzes are rigorously reviewed, monitored and continuously updated by our expert board to maintain accuracy, relevance, and timeliness.

  • Current Version
  • Mar 22, 2023
    Quiz Edited by
    ProProfs Editorial Team
  • Apr 27, 2017
    Quiz Created by
    Sumit Srivastava

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